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3 years ago

How much energy does an electric hair dryer use if it draws 8.3 amps of current when using 120 volts for 5.0 minutes?

Chemistry

2 answers:

Sloan [31]3 years ago

8 0

Answer: The energy drawn by electric hair dryer is 2,98,800 J

Explanation:

To calculate the power of the electric hair dryer, we use the equation:

$P=I\times V$

where,

P = power of electric hair dryer

I = Current of electric hair dryer = 8.3 A

V = voltage of electric hair dryer = 120 V

Putting values in above equation, we get:

$P=8.3\times 120=996W$

To calculate the energy of electric hair dryer, we use the equation:

$E=P\times t$

where,

E = energy of electric hair dryer = ?

P = Power of electric hair dryer = 996 W

t = time taken = 5.0 min = 300 s (Conversion factor: 1 min = 60 sec)

Putting values in above equation, we get:

$E=996\times 300=2,98,800J$

Hence, the energy drawn by electric hair dryer is 2,98,800 J

nalin [4]3 years ago

3 0

0.083kwh is the answer your welcome

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Read 2 more answers

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

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